/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 For what values of \(r\) does th... [FREE SOLUTION] | 91Ó°ÊÓ

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For what values of \(r\) does the sequence \(\left\\{r^{n}\right\\}\) converge? Diverge?

Short Answer

Expert verified
Answer: The sequence \(\left\{r^n\right\}\) converges for \(r = 0, -1 < r < 1\) and diverges for \(r < -1\) and \(r > 1\).

Step by step solution

01

Consider the cases for \(r\)

We can consider three main cases: Case 1: \(r = 0\) Case 2: \(r > 0\) Case 3: \(r < 0\)
02

Analyze convergence for \(r = 0\)

When \(r = 0, r^n = 0^n\). The sequence becomes \(\left\{0, 0, 0, \dots \right\}\). Since all the terms in the sequence are equal to 0, the sequence converges to 0. Hence, for \(r = 0\), the sequence converges.
03

Analyze convergence for \(r > 0\)

When \(r > 0, r^n\) is a geometric sequence with a positive common ratio. For \(0 < r < 1\), the terms \(\left\{r^n\right\}\) decrease as \(n\) increases, and approach 0. Therefore, the sequence converges, in this case, to 0. However, if \(r > 1\), the terms of the sequence \(\left\{r^n\right\}\) increase as \(n\) increases and grow without any bound. Thus, the sequence diverges for \(r > 1\).
04

Analyze convergence for \(r < 0\)

When \(r < 0\), the sequence becomes an alternating sequence. For example, if \(r = -2\), the sequence is \(\left\{-2, 4, -8, 16, \dots\right\}\). If \(-1 < r < 0\), the terms of the sequence \(\left\{r^n\right\}\) will be alternating but decreasing in magnitude and approach 0. Hence, in this case, the sequence converges to 0. On the other hand, if \(r < -1\), the terms of the sequence will still alternate, but their absolute values will increase without any bound. Therefore, the sequence will diverge for \(r < -1\).
05

Summaries the results

From the above analysis, we can conclude: 1. The sequence \(\left\{r^n\right\}\) converges for the following \(r\) values: \(r = 0, -1 < r < 1\). 2. The sequence \(\left\{r^n\right\}\) diverges for the following \(r\) values: \(r < -1\) and \(r > 1\).

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